Answer:
Circumference : 18.84 inches
Radius: 3 inches
-1.75, negative numbers go on the left of the negative sign going up (down) and since -1.75 is actually a greater number our answer is -1.75
Answer:dd and subtract the remaining numbers in the math problem. The sum will be the result of adding numbers, while the difference will be the result of subtracting them. For instance, in the math problem 4 + 3 - 5, the sum of 4 and 3 will be 7, and the difference between 7 and 5 will be 2.
Step-by-step explanation:
Answer:
Step-by-step explanation:
1.
To write the form of the partial fraction decomposition of the rational expression:
We have:

2.
Using partial fraction decomposition to find the definite integral of:

By using the long division method; we have:


<u> </u>

So;

By using partial fraction decomposition:


x + 20 = A(x + 2) + B(x - 10)
x + 20 = (A + B)x + (2A - 10B)
Now; we have to relate like terms on both sides; we have:
A + B = 1 ; 2A - 10 B = 20
By solvong the expressions above; we have:

Now;

Thus;

Now; the integral is:


3. Due to the fact that the maximum words this text box can contain are 5000 words, we decided to write the solution for question 3 and upload it in an image format.
Please check to the attached image below for the solution to question number 3.